18-03-09 The other root?

1) (a+√b)ⁿ = p+√q .................[ i]

let (a-√b)ⁿ = m+√n .................[ii]

multiplying both,

(a² -b)ⁿ = pm + p√n + m√q +√qn

since (a²-b)n is rational...

next..

adding [ i] and [ii]
(a+√b)n + (a-√b)n = 2(C₀aⁿ + C₂a^n-2b² +....)
= rational

=> √q+√n = 0

from the two bold equations...

-√q = √n

p(-√q) +m√q +pm -q = (a² -b)ⁿ

m√q - p√q =0 =>m=n

thus we get:

m+√n = p-√q

which question [7]

#17 question..

we can write the expression as:

√a+√a+x = x

=> √a+√a+√a+......√a+x = x

=> a + x = x²

=> x²-x-a =0

=> x = (-1 ± √1+4a)/2

one root is given... the other one is the other one......

well for the formal proof....

Let a+√b be the root of f(x)

now on dividing f(x) by p(x) we get Q(x) as quatient and λx+μ as remainder...

now let p(x)=(x-a+√b)(x-a-√b)=((x-a)²-b) ...(i)

now we can write f(x) = Q(x).p(x)+λx+μ

Putting a+√b in above identity
f(a+√b)=Q(a+√b).p(a+√b)+λ(a+√b)+μ

0=0+λa+λ√b+μ
(λa+μ)+λ√b=0 ...(ii)

so λ=0 hence μ=0 ...(comparing rational and irrational parts..)

so f(x)=Q(x).p(x)
p(x) is factor of f(x)

hence
a-√b is also root of f(x)

The above proof is true for any eqn having root of the form..a+(b)^1/n.. where n=2/I...and b≠0..(See the post below...)

Note this holds only for the below mentioned conditions...... not other fractional powers.... because for other powers...

as p(x)=(x-a+(b)^1/n)(x-a-(b)^1/n)=((x-a)²-(b)^2/n) so 2/n must be integer...............................(a)
or n must be 2/I... I is some integer...
so for n like 2,2/2,2/3,2/4....,2/(-1),2/(-2).... above rule holds....

so for other n(i.e.) n≠2/I (I is integer..)

for other n, problem arises in step(i).. as p(x) has irrational terms..........................from(a)
so remainder λx+μ will have irrational terms...(i.e.λ and μ will have irrational terms)hence in last step (ii) we can't say λ=0 and μ=0.. (as we can't compare rational and irrational parts then...)..

so a+(b)^1/n being a root of f(x) having rational coefficients
doesn't imply a-(b)^1/n to be root...for n≠2/I ,I is integer

For question in post 17
x=\sqrt{a+\sqrt{a+x}} , a\in (-\frac{1}{4},0)

If one root is \frac{1+\sqrt{4a+1}}{2} then other root is...
_________________________________

\frac{1-\sqrt{4a+1}}{2}
_________________________________________________

is also a root of equation... but when it is written in polynomial form... (as proof if for polynomial function..)

But in writing that we have to square... (twice) so case gain...

But we see in original equation x≥0

But for \frac{1-\sqrt{4a+1}}{2}≥0

\sqrt{4a+1}≥1

4a+1≥1
so 4a≥0 or a≥0 but a<0 so \frac{1-\sqrt{4a+1}}{2} is not a root of this equation..

Now dividing the polynomial form of given equation...
x^{4}-2ax^{2}-x+a^{2}-a
by x^{2}-x-a

obtained since two roots are known of this polynomial equation

we get x^{2}+x+(1-a)

which has \sqrt{4a-3} as discriminant... and for it to be positive a≥3/4.. but 1/4<a<0
so only two real roots..

So Is the above proof correct [7]
agar upar ka sab proof sahi hai tab....

This doesn't depend on degree but n..
and n in 1/n is not degree of equation... because....

consider equation.. x³+x²-x=0
here n is 2 in.... a+(b)^1/n not 3(which is degree of equation so true for odd degrees also.. not only for even degree..)... so n=2 therefore... other root will be a-(b)^1/n..( n=2 here)

so finally it is valid for square roots only...

utna n=2/I.. bla.. bla.. kiye... and finally only square root...
..(khoda pahad nikla chooha...)
as a+(b)^I/2=a+(b^I)^1/2.... :D :D

@priyam..... u wrote:

But we see in original equation x≥0

But for (1-√1+4a)/2 ≥0

=> √1+4a ≥ 1

how[7] [7] [7]

(1-√1+4a)/2 ≥0 => (1-√1+4a) >= 0

=> √1+4a <= 1

=> 1+4a <= 1

=> 4a <=0 !!

tum aur kuchh sochke woh likhe the kya [7]

skygirl

#36 Posted 10:04pm 18-03-09
Re: 18-03-09 The other root?

BUT I FEEL ^^^^ is quite perfect!!!

Priyam then y the contradiction??

Although this is nishant sir's thread, i will take the liberty to give two small hints:

(1) If a,b \in \mathbb{Q}, \sqrt{b} \notin \mathbb{Q} and (a+\sqrt{b})^n = p+\sqrt q, what is (a-\sqrt{b})^n in terms of p and q?

(2) Now, if p + \sqrt q = 0 what can be concluded about p and q?

(2) p=q=0

and if √q has a negative value and q is a perfect square..

then p≠0

(2) for this to be true p<0and equal to the magnitude of p =√ q
ya fir donno 0

any calc mistake [12]

subhaanallah... now d question got its flavour... [1]

awesome to read n ask... difficult to solve...!!! [2] [3]

post 36 wala proof sahi nahi hai kya.. :(

arey nahi.......

i told .. LET I is not an integer...

woh tum yeh proof kiye naa ... ki sirf integer ke liye hi conjugate roots hoga...

toh isiliye main yeh boli... isnt it obvious? ... now see post #43..

[9] [9] [9]

achha hum woh wala nahi padhe :P

yup!

but y shouldnt i ??

did i tell you i wont edit any of my posts ever ?? [3]

exactly prove kya karna hai aur.... merko bhi samajh ni aya....
[2]